luyỆn thỊ ĐẠi hỌc hÓa 2013

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Tr ng THPT chuyn Ng uyn Quang Diu

1 T hoa h c

LUYN THI AI HOC2013

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2 T hoa h c

Cu 1: t chy hon ton 0,1 mol hp cht X c cng thc 2( ) n HOOC CH COOH , cho snphm chy vo bnh nc vi trong thu c 30 gam kt ta. Y l mt ru no n chc khi b unnng vi 2 4 H SO c th to ra olefin. t chy hon ton mt este a chc to bi X v Y c t l

khi lng 2 2:CO H O tng ng l 176 : 63. Vy n c gi tr bng:

A. 2.

B. 0.

C.1.

D. 3.

Li giai:

Gi s C trong ru bng m . T d kin bi ta c

12 2

2 6

nm n m

n m

A

Cu 2: A l dung dch NaOH 1 M v KOH 3 M . B l dung dch HCl c 0 pH . Thm vo 200ml

dung dch B 1( )m g 2 5P O thu c dung dch C. Bit 100ml dung dch A phn ng hon ton vi C

thu c dung dch D . Chia D thnh 2 phn bng nhau.

Phn 1: em c cn thu c 17,8 (g) mui.

Phn 2: Tc dng vi Barinitrat d thu c 2 ( )m g kt ta. Bit mui photphat v hidrophotphat ca

Bari khng tan. Gi tr ca 1m v 2m ln lt l:

A. 1 10,65( )m g v 2 5,825( )m g .

B. 1 11,36( )m g v 2 5,825( )m g .

C. 1 10,65( )m g v 2 6,735( )m g .

D. 1 11,36( )m g v 2 6,735( )m g .

Li giai:

Ta c2

0, 4( ) H OOH n mol n khi p dng L bo ton khi lng ta c:

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3 T hoa h c

3 4 3 4 3 4 10,1.40 0,3.56 0,2.36,5 17,8.2 0,4.18 14,7 0,15( ) 11,65 H PO H PO H POm m g n mol m

Khi phn ng vi HCl tn ht 0,2(mol) OH v cn 0,2 mol cho 3 4 H PO xt t l th c cc ion l

2 4 H PO v24 : 0, 05( ) HPO mol

Khi 2 5,825m g

Cu 3:

un hn hp X gm 0,4 mol ancol etylic;0,3 mol ancol propylic;0,2 mol ancol butylic vi 2 4 H SO cnng.Sau khi tt c cc ancol u b kh nc.Lng olefin sinh ra lm mt mu va ng 0,2 mol dungdch 4KMnO .Cc phn ng xy ra hon ton.Khi lng nc to thnh trong s kh nc trn l:

.5, 4( ) A g

.9,9( ) B g

.10,8( )C g

.8,1( ) D g

Li giai

4.2 3. 0,3( )anken KMnO ankenn n n mol (nh lut bo ton e)

Suy ra ancoln tch nc to anken l 0,3( )mol cn li l to ete l 0,6( )mol do 2

0,6( ) 10,8 H On mol m g

Cu 4:

Sc ( )V l kh 2CO (ktc) vo 100ml dung dch 2( ) 1Ca OH M v 2,5 NaOH M thu c ( )a g kt

ta. Lc bkt ta sau sc tip 1,5 ( )V l kh 2CO (ktc) vo. Chcho phn ng xy ra hon ton r i

cho 2 BaCl d vo dung dch th thy c 2,7125 ( )a g kt ta. ( )V l kh 2CO trn c iu ch t

phn ng gia C v 3 HNO c. Khi lng 3 HNO cn dng l:

A. 20,16( )m g .

B. 23,03( )m g .

C. C A v B ng.

D. C A v B sai.

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4 T hoa h c

Li giai:

TN1:

Khi cho V lt 2CO v dung d ch ta c:

22 3 22CO OH CO H O

x (mol) 2x (mol) x(mol)

2 23 3Ca CO CaCO

x (mol) x(mol) x (mol) (

TN2:

Khi cho 1,5V lt2

CO vo :

22 32CO OH CO

0,225-x 0,45-2x 0,225-x

22 3 2 32CO CO H O HCO

2,5x-0,225 2,5x-0,225

2 23 3Ca CO CaCO

0,1-x (mol) 0,1-x (mol) 0,1-x (mol)

2 23 3 Ba CO CaCO

0,35 -2x (mol) --> 0,35-2x (mol)

Ti y ta gi i HPT

100

(0,35 2,5 ).197 (0,1 ).100 2,7125

x a

x x a

Suy ra 0,0914 x

Mt khc 3 2 24 4 2C HNO CO NO HOH

323,0 HNOm (gam)

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5 T hoa h c

TN3: Khi cho V lt 2CO v dung d ch ta c:

22 32CO OH CO HOH

x (mol) 2x (mol) x(mol)

2 23 3Ca CO CaCO

x (mol) x(mol) x (mol)

TN4: Khi cho 1,5V lt 2CO vo : 2CO ht, OH vn cn d :

22 32CO OH CO

1,5x (mol) 1,5x (mol)

3

3

:0,1( )

: 2,5 0,1

CaCO xmol

BaCO x

100(0,1 ) 197(2,5 0,1) 2,7125.100 0,08 x x x x

20,16m (gam)

Cu 5:

X l hidrocacbon mch h cha khng qu 3 lin kt trong phn t. Hn hp Y gm X v lng

2 H gp i lng cn thit hidro ha hon ton X . Cho hn hp Y qua Ni nung nng, sau phn

ng thu c Z c t khi hi so vi 2 H l313

. t ( )m g Z cn va 13,44(l) 2CO , hp thhon

ton sn phm chy vo 400ml dung dch 2( ) Ba OH 0,5 M v KOH 0,25 M thu c khi lng ktta l:

A.33,49g.

B. 35,46g.

C. 37,43g..

D. 39,4g.

Li giai:

2 2 2 2 2 2n n k n nC H kH C H

( )....... ( )....... ( )a mol ka mol a mol

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6 T hoa h c

Hn hp Y 2 2 2

2

:( )

: 2n n k C H a mol H ka

Hn hp Z 2 22

:( )

:n nC H a mol H ka

Ta c: 62 (14 2) 2 56( 1) 423tbZ

n a ka M k n

a ka

Tha mn2

4

k

nHn hp 4 6

2

:( )

: 4

C H aY mol

H a

t chy Z cng nh t chy Y :

4 6 2 2

114 3

2C H CO H O

11....... .......4 ( )

2a a a mol

2 2 2

12

H O H O

4 ...2 ( )a a mol

2 2

110, 6 2 0, 08 0, 08.4 0,32( )

2O COn a a a n mol

Ta c: 0,5( )OH n mol ; 2 0,2( ) Ban mol

23

0,5 0,5(0,32 ) 0,18

2 2COn < 2 Ban

Khi lng kt ta l 0,18.197 35, 46( )m g

B

Cu 6: Cho m gam K cho tc dng vi 500ml dd 3 HNO thu c dung dch M v thot ra 0,336lit

hn hp kh N gm 2 kh X v Y .Cho thm vo M dung dch KOH d th thot ra 0,224l khY .Bitqu trnh kh 3 HNO ch to ra 1 sn phm kh duy nht. m c gi tr l:

A. 3,12

B. 7,8

C. 12,48

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7 T hoa h c

D. 6,63

Li gi i:

Ta c: ddM ddKOH kh Y .Vy dd M cha 4 3 NH NO

3 3 4 3 28 10 8 3K HNO KNO NH NO H O

( )........................... ( )8 x

x mol mol

2 2

12

K H O KOH H

( )...... ( ).... ( )2 y

y mol y mol mol

4 3 3 3 2 NH NO KOH NH KNO H O

( ).......... ( ) y mol y mol

hh 2

3

1( ) : 1 0,336

( ) 0,015 0,01( )22 22, 4( ) :

X H y N mol y y y mol

Y NH y

4 3( ) 0, 01 0, 01 0,16( )

8 NH NO Y x

n du n x mol

Vy 39.(0,16 0,01) 6,63( )m g

D

Cu 7:

un 0,1mol hn hp X gm hai ancol no,n chc, mch h, ng ng k tip A v B ( A B M M )

vi 2 4 H SO 140o C thu c 1, 43( )g hn hp ba ete vi hiu sut 60% tnh trn A v 40% tnh

trn B .Thnh phn phn trm khi lng ca A trong hn hp X l:

.41,03% A

.64,88% B

.31,68%C

.51,06% D

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8 T hoa h c

Li giai:

Gi s

+,TH1: C 2 ancol u phn ng tch nc vi 40 H %,ta suy ra:

1, 43.2 71,50,1.0, 4hh

M (1)

+,TH2: C 2 ancol u phm ng tch nc vi 60 H 5 , ta suy ra:

1, 43.247,66

0,1.0,6hh M (2)

T (1) v (2) ta suy ra 2 ancol l 3 2 5, A CH OH B C H OH

Ti y ta gi3 2 5

( ), ( )CH OH C H OH

n x mol n y mol ( s mol ban u)

Ta c HPT:

40 60100 100

0,1

x y

x y suy ra x=0,06; y=0,04 . %A=50,06

Cu 8:

Cho 47 gam hn hp 2 ancol i qua 2 3 Al O un nng thu c hn hp Y gm 3 ete , 0,27 mol hai

olefin , 0,33mol hai ancol d v 0,42 mol nc .Bit r ng hiu sut tch nc i vi mi ancol u nhnhau v s mol cc ete u bng nhau .Ancol c khi lng phn t ln hn l:

A. 3 7C H OH

B. 4 9C H OH

C. 5 11C H OH

D. 3 5C H OH

Li giai:

ddng nhn thy hai ancol no n khng c 3CH OH

0,42 0, 270,05

3eten

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9 T hoa h c

0,27 0,15.2 0,33 0,9 X n

4709 X

M

=>c 2 5C H OH

gi2 5C H OH

n x

ROH n y

0,9 x y

46 ( 17) 47 x R y

5,629 y R

ta c

hiu sut to anken l 0,27 30%0,9

ta c

0,15 + 0,3y < y < 0,9 - (0,15 + 0,3x)

0,15< 0,7y < 0,48

=> 37,17< R 3 7C H OH

Cu 9: Thy phn ht 1 lng pentapeptit trong mi trng axit thu c 32,88 gam Ala Gly Ala Gly ; 10,85 gam Ala Gly Ala ; 16,24 gam Ala Gly Gly ; 26,28 gam Ala Gly ; 8,9 gam Alanin cn li l Gly Gly v Glyxin . T l smol Gly Gly : Glyxin l 5 : 4.

Tng khi lng Gly Gly v Glyxin trong hn hp sn phm l : A. 43,20 gam

B. 32,40 gamC. 19,44 gamD. 28,80 gam

L i gia i:

0,12 Ala Gly Ala Glyn mol; 0,05 Ala Gly Alan mol

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10 T hoa h c

0,08 Ala Gly Glyn mol; 0,18 Ala Glyn mol; 0,1 Alaninn mol

Pentapeptit l Ala Gly Ala Gly Gly

0,5. 0,5.(2.0,12 2.0,05 0,08 0,18 0,1) 0,35Pentapeptit Alan n mol

3. 1,05Gly Pentapeptit n n mol

Sau thy phn cn lai x mol Gly Gly v y mol Glyxin

Ta c h 2 1, 05 2.0,12 0, 05 2.0, 08 0,18 0, 42 0,15

4 5 0,

2{

1

x y x

x y y

.

Tng khi lng Gly Gly v Glyxin trong hn hp sn phm

132.0,15 75.0,12 28,8m g p n D

Cu 10 Thc hin phn ng crackinh m gam isobutan, thu c hn hp A gm cc hidrocacbon .Dnhn hp A qua bnh ng nc brom c ha tan 11,2 gam. Nc brom mt mu ht ,c 7,392 lit hm

hp kh B (ktc) gm cc hidrocacbon thot ra.T khi hi B so vi hidro bng177

. Gi tr ca m l:

A. 13,97 gamB. 8,46 gamC. 25 gam

D. 14,11 gamLi giai:

11,2 7,392 117.42 . .2 13,97( )

160 22,4 7m g A

Cu 11:

Cho 24,64 lit hn hp X (27.30 C , 1atm) gm 3 hidrocacbon. t chy hon ton hn hp X cho snphm chy cho hpf thhon ton vo dung d ch 2( )Ca OH d thy khi lng bnh tng 98,6 gam

.Cc hidro cc bon trong hn hp X thuc loi : A. parafin

B. olefin

C. iolefin

D. ankin

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11 T hoa h c

Li giai:

Cu ny ta th cho nhanh.

1 X n

t2 2;CO H O x n y n

nu A ta c: 44 18 98,6; 1 x y x y Gii h thy ng chn A

Nu l B : (44 18) 98,6 x tng ng 1,59 x n Suy ra loi v anken khng c cht no c sC nh hn 2.

Nu C hoc D : 44 18 98,6; 1 x y x y Tm ra snguyn t C trung bnh khng hp l nn cngloi.

Cu 12:

Ho tan ht 35,84g hn hp 2 3,Fe Fe O bng dung dch 31 HNO M ti thiu thu c dung dch A trong

smol 3 2( )Fe NO bng 4,2 ln 3 3( )Fe NO v ( )V l NO (ktc).Tnh thtch ca 3 HNO .

A.1,24 lt

B. 1,5 lt

C. 1,6 lt

D.1,8 lt

Li giai:

Fen a

2 3Fe On b

Ta c

56 160 35, 84a b

2 3 3 32 ( )Fe O Fe NO

b...................2b

3 32 ( ) 3 ( 3)2Fe Fe NO Fe NO

b..........2b.....................3b

3Fe Fe

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12 T hoa h c

a-b............a-b

4,2( ) 3a b b

=> 0,24a

0,14b

=>3 2( )

0,42Fe NOn

3 3( )0,1Fe NOn

0,24.2 0,180,1

3 NOn

30,42.2 0,1.3 0,1 1,24 HNOn

Cu 13:

t chy hon ton hn hp gm 3 3; ; x y x yC H COOH C H COOCH CH OH thu c 2,688 lt 2CO v

1,8 gam 2 H O . Mt khc cho 2,76 gam hn hp phn ng vi 30 ml dung dch NaOH 1 M , thu c

0,96 gam ancol. Cng thc ca x yC H COOH l:

2 5. AC H COOH

3. B CH COOH

2 3.C H COOH

3 5. DC H COOH

Li giai:

T smol ca 2 2,CO H O cng vi nhn xt smol ancol bng smol axit ta quy ngay v

2 2 20,03( ) n nmol C H O v 2( )a mol H O

Ta c 0,03 0,12 4n n .Chn ngay C Cu 14:

Mt bnh kn cha 0,08 mol axetilen 0,04 mol etilen; 0,08 mol vinylaxetilen; 0,23 mol 2 H .v mt t bt

Ni (th tch khng ng k). Nung bnh nhit cao n khi phn ng xy ra hon ton thu c hnhp Y .c t khi hi i vi 2 H l 19,55. Dn hn hp Y i qua bnh ng mt lng d dung dch

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13 T hoa h c

3 3 / AgNO NH thu c kt ta v 3,584 lt hn hp kh Z (ktc) thot ra khi bnh. Bit t khi hi ca

Z i vi 2 H l 19,5625.Th tch dung dch 2 Br 0,5M nhnht cn dng lm no hon ton hn hp

Z l:

.220( ) A ml

.180( ) B ml

.240( )C ml

.160( ) D ml

Li giai:

Ta c: 7.82hhm gam

7,82 0.239,1s

n mol

20,23

pu H n mol

2 H phn ng ht.

0,16 Z n mol

6,26 Z m gam

t2 2C H

n x mol v4 4C H

n y mol

26 52 7, 82 6, 26 1, 56 x y (1)

Ta c: x+y=0,2-0,16=0,04 (2)

T (1) v (2) ta c: x=y=0,02

dun =0,44-0,23-0,1=0,11=

2 Br n

2 0,22 Br V lit =220 ml

Cu 15:

Tin hnh ba th nghim sau vi hn hp X gm Al v kim loi M c ha tr khng i.

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14 T hoa h c

Th nghim 1:Ha tan hon ton a mol Al v b mol M trong dung dch 2 4 H SO c nng d thu

c 0,4 mol 2SO

Th nghim 2:Ha tan hon ton a mol Al v 3b mol M trong dung dch 2 4 H SO c nng d th

lng mui thu c tng 32gam so vi th nghim 1Th nghim 3:Ha tan hon ton 0,5a mol Al v b mol M trong dung dch 2 4 H SO c nng d thu

c 0,25 mol . 2SO l sn phm kh duy nht ca6S .Kim loi M l:

A. Mg

B. Zn

C. Ag

D.Cu

Li giai:

TN1: Bo ton smol e : 3a + nb = 0,8

TN2: Khi lng mui tng 32 gam do 2b mol M sinh ra b mol 2 4( ) n M SO ==> b*(2M + 96n) = 32

TN3: Bo ton smol e : 1,5a + nb = 0,5

==> 1,5a = 0,3 ==> a = 0,2 => nb = 0,2

Phn t lng mui = 322M 96n 160n M 32nb

vi n = 2 ==> M = 64 l Cu ===> cu D

Cu 16:

X l hidrocacbon mch h cha khng qu 3 lin kt pi trong phn t.Hn hp Y gm X v lng

2 H gp i lng cn thit hidro ha hon ton X .Cho hn hp Y qua Ni nung nng sau phn

ng thu c Z c tkhi so vi 2 H l313

.t m gam Z cn va 13,44 lit 2CO ,hp thhon

ton sn phm chy vo 400ml dung dch 2( ) Ba OH 0,5 M v KOH 0,25 M thu c khi lng ktta l:

A.33,49 gam

B.35,46 gam

C.37,43 gam

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15 T hoa h c

D.39,4 gam

Li giai:

hh X gm 2 2 2n n k C H 1 mol v 2 H 2k mol ===> khi lng X = khi lng Y M 4k ( M .lphn t lng X )

hhY gm 2 2n nC H 1 mol v 2 H d k mol ===> mol Y = 1 + k

Khi lng Y M 4k = 2.31(1 ). 3 62 503

k M k

Vi k = 2 ==> M = 54 ==> X l 4 6C H

==> Z gm 4 1 0C H 1 mol v 2 H 2 mol ==> t lmol 1 : 2

4 10 2 2 26,5 4 5C H O CO H O

a---------------6,5a--------4a

2 2 20,5 5 H O H O

2a---------a

Smol 2O = 7,6a = 0,6 ==> a = 0,08 ===> mol 2CO = 0,32

trong 400ml dung dch baz c 2 Ba = 0,2 , OH = 0,5 v K 0,1

mol 23CO = mol OH - mol 2CO = 0,5 - 0,32 = 0,18 < mol2 Ba = 0,2

===>khi lng kt ta 3 BaCO = 197x0,18 = 35,46 ===> cu B

Cu 17: Hn hp Z gm hai cht ng phn X v Y (mi cht cha mt loi nhm chc). Khi tchy z mol Z , ch thu c 2CO v 2 H O r i cho hp thsn phm chy vo bnh ng dung dchxt c d, khi lng bnh xt tng 328z gam. Nu cho tip nc vi d vo bnh xt th thu c 500z

gam kt ta. x mol X tc dng c Na to2

x mol 2 H , cn Y th khng. X c thc bao nhiu

cng thc cu to? A. 6

B. 7

C. 8

D. 9

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16 T hoa h c

Li giai:

Cng thc Z : n m t C H O z mol ---> nz mol 2CO v 0,5mz mol 2 H O

khi lng bnh NaOH tng = 44*nz + 9*mz = 328z ==> 44n + 9m = 328 (1)

2 2 3 22CO NaOH Na CO H O

nz----------------------------nz

2 3 2 3( ) 2 Na CO Ca OH CaCO NaOH

nz----------------------------------nz

smol kt ta = nz = 500100

z = 5z ===> n = 5 v t (1) ==> m = 12

Z l hp cht no c cng thc 5 12 t C H O

mol X : mol 2 H = 1 : 0,5 ==>X l ru n no, v Y l ete n no

S ng phn ancol n chc no 2 2n nC H O =22n vi (n< 6)

vi n = 5 ==>c 8 ng phn cu to ==> cuC .

Cu 18:

Hn hp X gm anehyt Y , axit cacboxilic Z , este T . t chy hon ton 0,2 mol X cn 0,625 mol

2O , thu c 0,525 mol 2CO v 0,525 mol 2 H O . Cho Y trong 0,2 mol X tc dng vi dd 3 AgNO dtrong 3 NH , un nng sau phn ng thu c m gam Ag ( 100h %). Gi tr ln nht ca m l

A. 32,4g

B. 10,8g

C. 16,2g

D. 21,6g

Li giai:

X chy cho mol mol 2CO = mol 2 H O ==> X gm cc hp cht n no : andehitY 2n nC H O a mol,

axit Z 2 2m mC H O b mol v este T 2 2 p pC H O

mol hh = a + b + c = 0,2 (1)

Bo ton smol oxy cho phn ng chy : a + 2b + 2c + 2*0,625 = 2*0,525 + 1*0,525

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17 T hoa h c

==> a + 2b + 2c = 0,325 (2)

(1), (2) ===> a = 0,075 v b + c = 0,125

===>andehitY 2n nC H O 0,075 mol

Nu n = 1, Y l HCHO trng gng cho 4x0,075 mol Ag ==> Agm = 108x0,3 = 32,4 gam

Nu n > 1, Y trng gng cho 2x0,075 = 0,15 mol Ag

===>khi lng ln nht l 32,4 gam ==> cu A

Cu 19:

Cho m gam Fe phn ng va vi dung dch 3 HNO long; sau khi phn ng xy ra hon ton ; thu

c 4,48 lit hn hp kh X gm NO v 2 N O (dktc) c t khi so vi 2 H l 16,75. Dung dch Y thu

c sau phn ng cha 92,3 gam mui. Mt khc cho dung dch Y trn tc dng vi mt lng vadung dch NaOH ; c cn r i nung trong khng kh n khi lng khng i th thu c 110,175gam cht r n khan.Gi tr ca m l:

A.25,2 gam.

B.12,6 gam.

C.37,8 gam.

D.16,8gam.

Li giai:

20,15( ); 0,05( ) NO N On mol n mol .

3 2

3 3

4 3

( )

( )

( ) 2 3 8 0,15.3 0,05.8 0,3( )

( ) 180 242 80 92,3 0,15( ) 25, 2

80( ) 69(2 3 ) 110,175 0,025( )( )

Fe NO

Fe NO

NH NO

n x mol x y z x mol

n y mol x y z y mol m g

x y x y z z moln z mol

Cu 20:

Hn hp X gm 2 cht hu c , A B n chc. Khi phn tch nh lng A cng nh B u thu c

50% C ; 5,56% H v 44,44% O . Ly hn hp X phn ng va vi 100 gam dung dch NaOH 12%, c cn phn dung dch sau phn ng thu c phn hi trong hi nc c khi lng 88 gam,phn r n khan un vi vi ti xt c hn hp kh v hi c tkhi so vi oxi bng 0,75. CTCT ca

, A B l

Li giai:

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18 T hoa h c

D dng tm c cng thc n gin nht ca A, B l 3 4 2( ) nC H O .

Dung dch NaOH ban u c khi lng nc l 100.12100 88( )100 lucsau

m g m X gm cc

este n chc CTPT 3 4 2C H O . 24 M Nh vy trong kh v hi phi c mt cht c 24 M trong

cc CT ch c 2 2 HO CH CH COONa l tha.

Cht th hai c hai kh nng l 2 M tc l Natrifomat v 18 M tc l HO COONa ( loi v ch to ra hn hp hi).=>pcm.

Cu 21: X phng ho hon ton 1kg mt loi cht bo gm triolein v axit bo t do c ch saxit l 7cn 3,125 mol KOH .Khi lng x phng thu c l:

A. 1080,75 gB.1083 gC.733 gD 896,75 g

Li giai:

ch s axit= 7 => s gam phn ng trung ha = 7(g) => n = 7/56 =0,125 => n. =0,125=> n. phn ng to glixerol= 3,125 - 0,125 =3 =>n.glixerol = 1ta co s : 1kg CB + 3,125 mol =>m xa phong + 0,125 + 1 mol glixerol =>m

Cu 22 : Hh hi gm , 1 anken, 1 ankin co cung s C trong phn t, c t khi so vi = 7,8. Saukhi cho hh qua Ni un nong phan ng xay ra hoan toan thi thu c hh mi co ti kh i so vi hh ban

u la20

9. Cng thc phn t anken va ankin la:

A/ v B/ v C/ v D/ v

Li giai:

Gi s s mol hh ban u l 1 mol

10,4520

9

hhsaun (mol)

21 0,45 0,55 H pun mol

20.55 H hhn mol

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19 T hoa h c

S dng quy tc ng cho cho hh u ta c

2

. .

13,6 0,5515,6 0,45

H

H C H C

n

n M

. 26,72 H C M

ankin M 26,72 2ankin M

Ankin 2 2C H ; Anken: 2 4C H

A

Cu 23: cho lung kh CO i qua ng s cha a gam hn hp A gm 2 3,CuO Fe O un nng.Sau 1

thi gian,trongng s cn li b gam hn hp cht r n B .Hp thkh sinh ra bng dung dch

2 Ba OH d thu c x gam kt ta.Biu thc ca a theo b,x l:

A. 16197

xa b B. 16

197 x

a b

C. 0,09a b x D. 0,09a b x

Li giai:

Ta c:

2 16CO Oa b

n n

3 2 16 BaCO COa b

n n (do 2( ) Ba OH d)

3

197( )16 BaCOa b

m x

16197

xa b

p n B

Cu 24:

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20 T hoa h c

, , , X Y Z T l cc kh khc nhau trong scc kh: 3 2 2 2, , ,O F Cl O . Sc X vo dung dch KI + h tinh

bt,thy dung dch ho xanh.Sc Y vo dung dch NaCl sinh ra kh Z . Thi Z vo Ag h nng nh

khng thy Ag mt nh kim.Cho tn m vo T thy bng chy.Hi 3O l kh no?

.Li giai:

3 2 2 22 ( ) 2KI O X H O O KOH I

2 2 2( ) 2 4 ( )F Y H O HF O Z

2 ( )O Z Ag khng phn ng

2:T O

ap an A

Cu 25:

Cho m gam hn hp bt gm ,Fe Cu , 2 3Fe O tc dng va dung dch HCl (lng dung dch HCl

dng ti thiu ) thu c dung dch A gm 2FeCl v 2CuCl vi smol 2FeCl bng 9 ln smol

2CuCl v 5,6 lit 2 H (dktc)khng cn cht r n khng tan .C cn dung dch A thu c 127,8 gam cht

r n khan .m c gi tr : A.68,8 gam

B. 74,4 gam

C. 75,2 gam

C. 61,6 gam

Li giai:

2 20,9 . 0,1 .Fe Cun mol n mol

22

( 0 ) 0,25 H Fen d Fe H n mol .

3 2 22 2 ;Cu Fe Fe Cu

3 22Fe Fe Fe

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21 T hoa h c

Tnh c 2 ( ) 0,9 0,25 0,1.2 45Fen Fedusr mol .

0,15 . 0,4Fedu Fen mol n bd mol .

2 3

310, 25 .2Fe On nFe mol

m A

Cu 26:

in phn 200 ml dung dch gm 3 AgNO 0,1M v 3 2( )Cu NO 0,2 M vi in cc tr, cng dng

in 5 I A, trong thi gian 19 pht 18 giy. Khi lng dung dch sau in phn gim m gam. Tnh m, bit hiu sut in phn 80%, bqua s bay hi ca nc.

A. 3,92g B. 3,056 g C. 6,76g D. 3,44g

Li giai:

Hiu sut l 80% => 0,06.0,8 0,048en

Ghi phng trnh in phn

=> 0,02 Agn

0,048 0,020,014

2Cun

20,012On

0,02.108 0,014.64 0,012.32 3,44m

Cu 27:

Ho tan hon ton m gam hn hp gm Na v Al ( t l s mol l 2:1) vo nc d c dung dch X

-Cho t t 200ml dung dch HCl 1 M vo X th thu c t gam kt ta.

-Nu cho t t 300ml dung dch HCl 1 M vo X th thu c 1, 25 t gam kt ta.

Gi tr ca m l:

A. 8,76

B. 9,24

C. 12,6

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22 T hoa h c

D. 7,92

Li giai:

Ta c:

0,278

0,120,3 2 1, 253 78

t a

a Aa t a

Cu 28.

Ly 23,49(g) mt amin n chc mch h X ( X cha khng qu 3 lin kt trong phn t) tr n vi252 lt khng kh. Bt tia la in t chy ht X , hn hp sau phn ng a v iu kin tiu chunth thu c 235,368 lt kh.(Cc kh o iu kin tiu chun). S ng phn ca X l:

A. 8.

B. 16.C. 17.

D. 32.

Li giai:

Smol kh thu c = 10,5075mol

Smol khng kh = 11,25mol nn s mol nit = 9mol v smol oxi = 2,25mol

Do hn hp kh gm: 2 2 2; ;CO N O cn d

CTTQ ca amin n chc X: 2 3 2n n k C H N

=>s mol oxi p = nx x 0,5n 0,75 0,5k x 1,5n 0,75 0, 5k (mol)

=>s mol oxi d 2,25 x 1,5n 0,75 0,5k

=> nx 0,5x 2,25 x 1,5n 0,75 0,5k 1,5 075

=> x 2n 1 2 2,97

M: x 14n 17 k 23,49

=>5,4n 27 44,01k k 0 ; n 5

X l : 5 13C H N c 17 ng phn

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23 T hoa h c

Cu 29.

Khi ha hi 58,5 gam mt hydrocacon X th th tch hi bng th tch ca 12,6 gam nit (o cng iukin). X phn ng vi nc brom trong 4CCl theo t l1: 2 v cho phn ng vi 3 3 / AgNO NH . un

nng X vi dung dch 4KMnO ti khi ht mu tm, r i thm lng d dung dch HCl c vo hn hp

sau phn ng thu c axit p-phtalic ng thi gii phng kh 2CO v 2Cl . Xc nh CTCT ca X .

Li giai:

Ddng tm ra X l 10 10C H c 130 M u .

X c mt nhnh C CH v nhnh cn li l l etyl.

v tr para 6 4 2 5 p CH C C H C H

Cu 30 .

Ha tan 6,58 gam mt hp cht v cA vo 100 gam 2 H O thu c dung dch B cha 1 cht tan duy

nht. Cho mt lng mui 2 BaCl khan vo B thy to ra 4,66 gam kt ta tr ng, lc bkt ta thu

c dung dch C . Cho mt lng Zn d vo C thy thot ra 1,792 lt 2 H ktc. Xc nh CTPT

ca cht A .

Li giai: T bi ta c th suy ra mt sd kin nh sau:

(1) A tan tt trong nc (c thA l cht tc dng c vi 2 H O )

(2) A to c mi trng Axit hay Bazo (mnh) v ha tan c km

(3) A to kt ta vi 2 Ba nn A phi mang ion 23CO hay23SO hay

24SO (my ci ny thng dng

nn lit k trc)

(4) A l cht v c

Kt hp 2 v 3 ta thy va mang gc axit khng th cho mi trng bazo mnh c nn A cho mitrng axit nh...

M khi A cho mi trng axit (mnh) tc dng vi Zn (gii phng kh hidro) th ta loi lun 2 ion23CO v

23SO

TH1 : A l mui axit 4 HSO

Ta thy 4 HSO to kt ta vi 2 BaCl

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24 T hoa h c

40,02 BaSOn mol

20,08 0,16 H H n mol n mol

Mt s chnh lch khp khing!

Ta c th l gii v bi u c ni 2 BaCl d nn ta c thcho r ng 2 BaCl thiu !

Ta sda vo d kin (an ton) l Zn d

0,16 H

n mol tc l4 4

0.16 0,16 97 6,58( ) HSO HSO

n mol m g LO I !

V anion (cha c cation) m ln hnm ban u r i

TH2: Cht tan l 2 4 H SO

Ta cng da vo d kin chc chn l Zn d

20,08 0,16 H H n mol n mol

Vy2 4

0,08 H SOn mol

* Nu ha tan 2 4 H SO ta xem xem c ph hp khng

2 4

0,08 98 7,84 H SOm g

Vy cht A khng phi l Axit

* Lno l 3SO ?

3 2 4 30,08 6,4 ( 6,58 )SO H SO SOn n molm g g

A L OLEUM 2 4 3. H SO nSO (t smol l x)

Ta c:

x. 98 80n 6,58g (khi lng A )

x. n 1 0,08 mol (smol axit A tan ra m thnh)

GI I RA x=0,01 mol ; n = 7

Vy cng thc ca A l 2 4 3.7 H SO SO

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25 T hoa h c

Cu 31:

Cho 10,6 gam hn hp A gm mt kim loi kim X v 1 kim loi kim th Y tc dng ht vi 300mldung dch HCl 1M ( long), thu c dung dch B v 5,6 lt kh ( ktc). Dn t t kh 2CO vo dung

dch B . Sgam kt ta cc i to thnh:

A. 20gam

B. 15gam.

C. 10gam

D. 5 gam

Li giai:

0,3 HCln ; 2 0,25 H n HCl p ht v KL p vs 2 H O

Ta c: 0,25.2 0,5( )en mol

* A gm ton KL kim10,6

21, 20,5tb

M

*A gm ton KL kim th 10,6

42,40,25tb

M

Hn hp A gm Na v Ca vi smol ln lt l a ; b ta c h:

23 40 10,6 0,2 ( )2 0,5 0,15

a b a mola b b

Khi lng kt ta max l: 23

.100 0,1.100 10( )CO

m n gam C

Cu 32:

Thc hin phn ng este ho gia axit axetic(ly d) v hn hp gm 8,4 gam 3 ancol l ng ng caancol etylic. Sau phn ng thu c 16,8 gam 3 este. Ly sn phm ca phn ng este ho trn th chin phn ng x phng ho vi dung dch NaOH 4 M th thu c m gam mui: (Gi s hiu sutphn ng este ho l 100%). Gi tr cam :

A.10gam

B.16,4gam

C.8,2gam

D.12,3gam

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26 T hoa h c

Li giai:

Gi x l smol axit axetic phn ng,BTKL ta c:60 8, 4 16,8 18 0,2 x x x

3

0,2.82 16,4CH COONam B

Cu 33: pH ca mt dung dch yu l 2,34. Sau khi pha long gp 3 ln th 2,6 pH . Hng sphn liaxit ?

A. 46,25.10

B. 46,625.10

C. 46.10

D. 45,892.10

Li giai:

t nng ban u l C . Ta c2,34 2,34

2,34

10 1010

K C

(1)

Khi pha long gp 3 ln, nng sgim i 3 ln, ta c:2,6 2,6

2,6

10 10

103

K C

(2)

Cho (1) = (2) gii ra C , thay ngc li vo (1) ta tnh c K = 46,625.10

Cu 54:

Cu 34: X c CTPT . X tc dng vi dung dch (un nng nh) hocu c kh thot ra. Ly 18,60 gam X tc dng hon ton vi 400ml dung dch 1M. Sau

phn ng c cn dung dch r i nung nng cht r n n khi lng khng i th c m gam. Xc nhm ?

A. 22,75B. 19,9C. 20,35D. 21,20

Li giai:

Ta c cht l 3 3 2 3( )CH NH CO ...

Vy ta c phn ng xy ra l:

3 3 2 3 2 3 3 2( ) 2 2CH NH CO NaOH Na CO CH NH
http://latex.codecogs.com/gif.latex?\dpi{120}%20NaOHhttp://latex.codecogs.com/gif.latex?\dpi{120}%20HClhttp://latex.codecogs.com/gif.latex?\dpi{120}%20NaOHhttp://latex.codecogs.com/gif.latex?\dpi{120}%20C_3H_{12}O_3N_2

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27 T hoa h c

Vy2 3du NaOH Na CO

m m m =19,9 g

Cu 35:

Hn hp X gm , v mt s ancol no, n chc, mch h (trong

v c smol bng nhau). t chy hon ton 5,444 gam X r i hp th ton b sn

phm chy vo dung dch d thy khi lng bnh tng ln 16,58 gam v xut hin m gamkt ta. Xc nh m?

A. 42,158 gamB. 43,931 gamC. 47,477 gamD. 45,704 gam

Li giai:

t cng thc chung ca X l 2 2n nC H O

PTHH:

2 2 2 2 2

3( 1)

2n nn

C H O O nCO n H O

a ................................... a n ............ ( 1)a n

Theo bi ra:

(14 18) 5,444

44 18 ( 1) 16,58

a n

an a n

0,232

0,122

an

a

3 20,232( ) BaCO COn n mol

30,232.197 45,704 BaCOm g

D

Cu 36:

Hp cht hu c X a chc c cng thc phn t 9 14 6C H O . Thc hin phn ng x phng ha hon

ton X sn phm thu c l hn hp 2 mui ca 2 axit hu c n chc (trong c 1 axit c mchcacbon phn nhnh) v hp cht hu c a chc Y . em 13,08 gam X tham gia phn ng trng bcth khi lngAg ln nht thu c l
http://latex.codecogs.com/gif.latex?\dpi{120}%20Ba(OH)_2http://latex.codecogs.com/gif.latex?\dpi{120}%20C_2H_4(OH)_2http://latex.codecogs.com/gif.latex?\dpi{120}%20C_3H_8http://latex.codecogs.com/gif.latex?\dpi{120}%20C_2H_4(OH)_2http://latex.codecogs.com/gif.latex?\dpi{120}%20C_3H_8

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28 T hoa h c

A. 12,96 gam.

B. 25,92 gam.

C. 27 gam.

D. 6,48 gam.

Li giai:

Cng thc phn t 9 14 6C H O l este 3 chc no, tc dng vi NaOH cho 2 mui ca axit hu c n

chc 1 2 R COOH R COOH trong 1 xit c mch carbon phn nhnh, v d gc 1 R phn nhnh==> gc 1 R c t nht 3 carbon ==> cng thc cu to thu gn ca este l:

1 2 2 3 5( )( ) R COO R COO C H

Este X trng gng ==> X c gc H COO

==> X l

3 2 2 3 5 3 2 3 5( ) )( ) 3 ( ) 2 (CH CH COO H COO C H NaOH CH CH COONa HCOONa C H OH

smol este = 13,08 0,06218

==> smol formiat 2.0,06 0,12

==>smol Ag = 2 mol formiat = 0,24 ==> khi lng Ag = 108.0,24 25,92 ===> cu B

Cu 37:

Cho hn hp X vo bnh r ng phn ng este ha xy ra vi hiu sut 80%. Cho t t bt Na vobnh n khi khng cn kh thot ra th thu c 2m gam mui. Bit trong X smol axit bng smol

ancol, gi tr ca 2m l:

A. 34,8 B. 3,76 C. 6,96 D. 13,36

Li giai:

y este c to ra skhng tc dng vi kim sinh ra t phn ng gia Na v 2 H O to muihu c v khng c iu kin un nng

Cch 1:

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29 T hoa h c

2

2

1,20,2

1

CO

axit ancol H O

nn n

n

Bo ton Oxi2 2

1,4 44,8O On m

Bo ton khi lng 26( ) 652

axit ancol X tb

M M m g M

T y ddng suy ra 2

2 2

72( )

58( )axit

ancol

M CH CH COOH

M CH CH CH OH

n y c thd dng tm c 2 6,96m

Cch 2:

Ta khng cn tm cng thc cu to, mui 2m l mui sinh ra t phn ng gia axit d v ancol d viNa

226 26.0,8 5,2 5,2 22.0,08 6,96 Xdum m

Hay bi ny tc gi l cho Na vo lc cha thc hin p 2 26 0,4.22 34,8m

Cu 38:

Ha tan ht 17,94 gam 1 kim loi kim vo 1 lng nc d thu c dung dch X . Cho dung dch X tc dng vi 36,92 gam

2 5P O th thu c dung dch Y ch cha 2 mui c nng mol bng nhau. Kim

loi kim l:

A. Na

B.Rb

C. K

D.Li

Li giai:

Dung dch Y ch cha 2 mui nn ch xy ra 2 trng hp sau:

2 424

: 0, 26( )(1)

: 0, 26( )

H PO mol

HPO molhoc

24

34

: 0, 26( )(2)

: 0, 26( )

HPO mol

PO mol

Xt TH(1) 0,26.3 0,78 23( )kl kln M Na A ,trng hp kia khi xt.

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30 T hoa h c

Cu 39:

Nung hn hp gm x mol 3 2( )Fe NO , y mol 2FeS v Z mol 3FeCO trong bnh kn cha khng kh d.Sau khi phn ng xy ra hon ton a bnh vnhit ban u thy p sut bnh khng di so vi banu. Mi lin hgia x;y;z:

A. 6x + 2z = y

B. 3x +z = y

C. 9x + 2z = 3y

D. 6x + 4z = 3y

Li giai:

2 3 2 3 2 2

12 ( ) 4

2Fe NO Fe O NO O

2 2 2 3 24 11 2 8FeS O Fe O SO

2 2 2 3 24 2 4FeCO O Fe O CO

Mun p sut khng i th s mol kh tng ln phi bng smol kh gim i,tc l:

112 2 3

4 4 4 x z y

x z y y x z B

Cu 40:

Cho hn hp X gm mt axit cacboxylic n chc, mt ancol n chc v este to bi axit v ancol. t chy hon ton 1,55 gam hn hp X thu c 1,736 lt 2CO ( ktc) v 1,26 gam 2 H O . Mt

khc khi cho 1,55 gam hn hp X tc dng va ht vi 125 ml dung dch NaOH 0,1 M to ra m gammui. Sau phn ng tng s gam ancol thu c l 0,74 gam v ng vi 0,01 mol. Vy gi tr ca m l:

A.1,175 gam

B.1,205 gam

C.1,275 gam

D. 1,305 gamLi giai:

Ta c:2

0,0775COn mol

2 00,07 0,03( ) H On mol n mol .

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31 T hoa h c

Gi smol axit ,ancol, este ln lt l , , ( )a b c mol ta c

0,01 0,0075

0,0125 0,005

2 2 0,03 0,005

b c a

a c b

a b c c

=>gc axit l 2 3 1,175C H m g

Cu 41:

Hai bnh kn A , B u c dung tch khng i V lt cha khng kh (21% oxi v 79% nit vth tch).Cho vo chai bnh nhng lng nh nhau hn Z n S v 2FeS . Trong bnh B cn thm mt t bt S

(khng d). Sau khi t chy ht hn hp sunfua v lu hunh, lc trong bnh A oxi chim 3,68% th tch, trong bnh B nit chim 83,16% th tch. % th tch ca 2SO trong bnh A l :

A.13,16%.B. 3,68%.

C. 83,16%.

D.21%.

Li giai:

Th tch sau p bnh A v B l nh nhau v lng 2 N bnh A v B bng nhau

2 / 100 3,68 83,16 13,16SO BV A

B i 42:

Nhit phn hon ton m gam hn hp X gm KCl , 3KClO , 2CaCl , 2CaOCl , 3 2( )Ca ClO thu c

cht r n Y v 2,24 lt kh 2O (ktc). Ha tan Y vo nc c dung dch Z . Cho dung dch Z tc dng

vi dung dch 2 3 Na CO d thu c 20 gam kt ta. Nu cho dung dch Z tc dng vi dung dch

3 3 / AgNO NH d th thu c 71,75 gam kt ta. Gi tr ca m l:

A.50,6 g.

B. 124,85 g.

C. 29,65 g.

D. 32,85 g.

Li giai:

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32 T hoa h c

2

0,2; 0,5 0,5 0,2.2 0,1CaCl AgCl KCln n n

2 2

32,85CaCl KCl Om m m m D

Cu 43:

Thy phn este A bng dung dch NaOH , thu c mui B v cht hu c D . Cho B phn ng vidd 3 3 / AgNO NH thu c Ag v dung dch X . Dung dch X va tc dng vi NaOH va tc dng

vi 2 4 H SO u sinh kh v c. Bit D c cng thc 2( )CH O n v tha mn s :

02 /

2( ) H t Ni HCl

n D F CH Cl

Khi t 0,1 mol A thy cn V lit 2O (kc). Gi tr V l:

A. 5,6

B.8,96

C.5,6 hoc 6,72

D.6,72 hoc 8,96

Li gi i:

T 2( ) nCH Cl tng ng 3( ) nCH bin lun 2n D l 2 4 2C H O

2 2 2 2 2 2( ) n D F CH Cl HOC CH OH HO CH CH OH Cl CH CH Cl

B cho phn ng trng gng B l mui HCOONa .

X c 2 trng hp:

22 2 HCOO CH CH OOC H NaOH H COONa HOC CH OH

2 2 2( ) H COO CH OH CH OH NaOH H COONa HOC CH OH H O

Nu X l

24 4 4 2 2 23 4 2 0,3 6,72O H COO CH CH OOC H C H O O CO H O n V

Nu X l

22 3 6 4 2 2 2( ) 2,5 3 3 0,25 5,6O H COO CH OH CH OH C H O O CO H O n V

Chn C .

Cu 44.

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33 T hoa h c

Hp thhon ton V lt 2CO vo dung dch 2( )Ca OH aM th thu c 1m gam kt ta. Cng hp th

( 3, 36)V lt 2CO vo dung dch 2( )Ca OH th thu c 2m gam kt ta. Bit1

2

32

mm

. Nu thm

1( )V V lt 2CO vo dung dch 2( )Ca OH trn th thu c lng kt ta cc i. Bit 1m bng3

7khi

lng kt ta cc i. Gi tr ca 1V l:

A.0.67lt

B.1.493lt

C.2.016 lt

D.1.008 lt

Li giai:

Coi2( )Ca OH

n a mol

Ta c : 122,4

V V a

1 322, 4 100 7

mV a (1)

23,36 2222,4 100 7

mV aa (2)

T (1) v (2) c 760

a

T tnh ra 1,493V lt

p n B

Cu 45:

Cho 3,6 gam Mg tc dng vi dung dch hn hp 2 4 H SO v 3 2( )Cu NO , un nh n khi phn ng xy

ra hon ton thu c 0,224 lt 2 N (ktc), dung dch A v 2,64 gam hn hp B gm hai kim loi c

cng smol. C cn cn thn dung dch A thu c m gam mui. Gi tr ca m l:

A.15,48

B.18,56

C.4,92

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34 T hoa h c

D.13,62

Li giai:

0,15 Mgn mol

20,01 N n mol

Cht r n thu c gm Cu v Mg d : 0,03duCu Mg

n n mol

0,12 pu Mg

n mol

bo ton e

2 2 Mg Mg e

0,12----0,12 -----0,245

22 10 N e N

0,02--0,1----0,01

2 2Cu e Cu

0,03-----0,06

5 38 N e N

0,01-----0,08-----0,01

30,02

du NOn mol

24

0,11SO

n mol

4 3 3?mu i Mg NH NO NO

m m m m + 24SO

m =4,92 gam

Cu 46:

Mc ni tip 2 bnh in phn, bnh mt cha 1 lt dung dch3

AgNO 1 M , bnh hai cha 300ml dung

dch 3FeCl 1 M . Tin hnh in phn 2 bnh vi cng dng in 1,93A trong thi gian t giy.Sauin phn, ly dung dch trong bnh mt cho vo bnh hai. Sau khi phn ng xy ra thu c V lt kh NO(ktc) duy nht v 105,85 gam kt ta, dung d ch sau phn ng cha mui ca hai kim loi( bit hiusut cc qu trnh l 100%). Gi tr ca t v V l:A. 8000 v 2,24B. 11882 v 1,33

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35 T hoa h c

C. 10000 v 3,36D. 8118 v 0,91

Li giai:

31,0 AgNOn mol ; 3 0,3FeCln mol

Bnh 1:

Ag e Ag

x--------x

2 22 1/ 2- 2 H O e O H

-----------x--------------------x

Bnh 2:

3 2Fe e Fe

x---------x--------x

21/ 2Cl e Cl

x--------x

Khi tr n 2 dung dch sau in phn vo nhau thy kh NO bay ra v xut hin kt ta. Nn:

+ Dung dch sau in phn bnh 1 cn Ag

+ Dung dch sau in phn bnh 2 c cha 2Fe

Hn thna H phi phn ng ht qu trnh kh 3 NO thnh NO .

Dung dch sau khi kt thc phn ng cha 2 mui kim loi l 3Fe v Ag

Ag Cl AgCl

(0,9-x)

2 33 23 4 3 2Fe H NO Fe NO H O

0,75x-------x

2 3Fe Ag Fe Ag

z----------------------------------z

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36 T hoa h c

x 0,75x z =>z 0,25x

143,5 0,9 x 27x 105,85 x 0,2mol

V 1,12 lt v t 10000s .

Cu 47:

Ha tan ht35,84 gam hn hp 2 3,Fe Fe O bng dd 31 HNO M ti thiu thu c ddA trong o s mol

ca 3 2( )Fe NO bng 4,2 ln 3 3( )Fe NO v ( )V l kh NO . Th tich dd 3 HNO :

A.1,24l

B.1,5l

C.1,6l

D.1,8l

Li giai:

Gi2 3

Fe

Fe O

n x

n y

2 3 3 3 3 26 2 ( ) 3Fe O HNO Fe NO H O

y ...........6y ..........2y

3 22 3Fe Fe Fe

y.......2y........3y

3 3 3 24 ( ) 2Fe HNO Fe NO NO H O

..4( )...... x y x y x y

Ta c h :

56 160 35,84

3 4,2( )

x y

y x y

3

0,241,24

0,14 HNO x

n y

(mol) A

Cu 48.

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37 T hoa h c

Hn hp X gm 2 este ca 2 axit k tip trong dy ng ng l 1 R COOR v 2 R COOR . t chy

hon ton 20,1 gam X cn 146,16 lt khng kh (ktc, cha 20% 2O v 80% 2 N ) thu c 46,2gam

2CO . Mt khc, nu cho 3,015 gam X tc dng va vi NaOH c 2,529 gam hn hp mui.Cng thc ca 2 este l.

A. HCOOH v 3CH COOH

B. 3 2 2CH COOCH CH CH v 2 5 2 2C H COOCH CH CH

C. 3 3CH COOCH v 2 5 3C H COOCH

D. 3 2CH COOCH CH v 2 5 2C H COOCH CH

Li giai:

TN 1:

2 2 21,05; 1,305 0,87CO O H On n n

2.1,05 0,87 1,305.20,18

2esten mol

este M =111,67

TN2:

1, 2 2,5290,027 67 26,670,027este R Rn M

3 2 5 3 5; ;CH C H C H

Cu 49.

Mt hn hp gm 3 cht thuc dy ng ng aren l: A ( 2 6n nC H ); B ( 2 6n nC H ); C ( 2 6m mC H )

vi n n m ; trong A v C c smol bng nhau v cch nhau k ch t trong dy ng ng. Khit x gam hn hp cn y gam 2O . Biu thc lin hx, y, m, k l.

A.24 3 48 1024 7 24 7

x y x ym k

x y x y

B. 48 3 48 10 224 7 24 7

x y x ym k

x y x y

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38 T hoa h c

C. 24 3 48 1048 7 24 7

x y x ym k

x y x y

D.12 3 24 10

224 7 24 7

x y x ym k

x y x y

Li giai:

Gi cng thc tng qut ca hn hp l 2 6a aC H vi a l s C trung bnh

2 6 21,5( 1)a aC H a O 2 2( 3)aCO a H O

ta c:

3214 6 1,5( 1)

y xa a

24 324 7

x ya

x y(1)

V A,C cch nhau k cht nn 1m n k

V n a m

nn 1m k a m

1a m a k

(2)Thay (2) vo (1) ta c

24 3 48 1024 7 24 7

x y x ym k

x y x y

Vy p n A

Cu 50:

Hp cht hu cY vng thm c phn trm khi lng ca C = 63,16% ; H = 5,26% v O = 31,58%.

Khi ha hi 18,24 gamY th th tch bng th tch ca 3,36 gam 2 N o cng iu kin. Bit r ng khi choY tc dng 3 NaHCO th smol 2CO sinh ra bng smol Y phn ng v tc dng NaOH theo tl

mol 1:1. Scng thc cu to ph hp ca Y l.

A. 4

B. 5

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39 T hoa h c

C. 6

D. 7

Li giai:

152Y M . T phn trm khi lng suy ra CTPT ca Y l 8 8 3C H O

Y tc dng 3 NaHCO th smol 2CO sinh ra bng smol Y phn ng nn Y c mt nhm COOH

Y tc dng NaOH theo t lmol 1:1 nn Y Khng c OH gn tr c tip vo vng thm

Y c th l 6 4 2COOH C H CH OH ( , , )o p m v 6 5( ) HOOC CH OH C H v

6 4 3COOH C H O CH ( , , )o p m

Vy c 7 CTCT .

Cu 51:

Hn hp A gm 2 cht hu c , ( , , ) X Y C H O ch cha mt loi nhm chc. Cho m gam A phn ngva 100 ml dung dch NaOH 1,5 M thu c hn hp mui ca axit hu c v phn hi c cha

mt cht hu c Z. Ly ton b lng Z sinh ra tc dng vi 3 3 / AgNO NH thu c 25,92 gam Ag.

un hn hp mui trn vi vi ti xt thuc 1 cht kh duy nht c tkhi so vi 2 H l 8. Bit r ng

, X Y hn km nhau 1 nguyn tcacbon trong phn t.Gi tr ca m l.

A. 11,88gam

B. 13,44gam

C. 11,67 gam

D. 10,20 gam

Li giai:

V un hn hp mui vi vi ti xt thu c 1 cht kh duy nht c tkhi so vi 2 H l 8

M kh ny l 4CH nn 2 gc axit l 3CH COO v 2OOC CH COO

0,15 NaOH n mol ; 0,24 Agn mol

Z l anehit n chc c 2C tr ln nn 0,12 Z n mol

VX ,Y hn km nhau 1 C trong phn t nn X l 3 2 3CH COO C H th Y l

2 HOOC CH COOH

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40 T hoa h c

0,12 X n mol;0,15 0,12

0,0152Y

n mol

Vy 11,88m gam

p n A Cu 52:

Mt cht hu c A gm 3 nguyn t , ,C H O c khi lng phn t nh hn 150dvC . t chy hon

ton 14,4 gam A thu c 33,6 gam hn hp 2CO v 2 H O trong oxi chim 76,19% vkhi lng.

Ly 21,6 gam A tc dng va vi dung dch NaOH thu c mt ancol v dung dch Y ch 1 mui. Axit ha ton bdung dch Y, sau cho tc dng vi 12,88 gam 2 5C H OH (xt 2 4 H SO ) thu c m gam este thu n chc. Bit hiu sut este ha l 65%. Gi tr m l.

A. 13,286gam

B. 20,440gam

C. 14,235gam

D. 18,615gam

Li giai:

33,6 gam hn hp 2CO v 2 H O trong oxi chim 76,19% vkhi lng nn 1,6On mol

Ta c2

2

CO

H O

n x

n y

44 18 33,6

2 1,6

x y

x y

0,6

0,4

x

y

Vy A c : : 3: 4 : 2C H On n n

M 150 A M nn A l 3 4 2C H O hoc 6 8 4C H O

V A tc dng vi dung dch NaOH thu c mt ancol dung dch Y ch 1 mui nn A khng th l

3 4 2C H O A l 6 8 4C H O (c k=3)

0,15axit n mol; 2 5 0,28C H OH n mol

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41 T hoa h c

TH1 : A l este mch h to bi axit no 2 chc v ancol n chc c 1 pi

A c th l

+) 2 3 5 HOOC COO CH C H th 0,65.(0,14.88 0,28.29) 13,286estem gam

+) 2 2 2 3 HOOC CH COO CH C H th 0,65.(0,14.102 0,28.29) 14,56estem gam

TH2 : A l este mch h to bi ancol no n chc v axit 2 chc c 1 ni iC C

Axit l 2 2n n HOOC C H COOH

+) n =2 th 0,65.(0,14.114 0,28.29) 15,652estem gam

+) n =3 th 0,65.(0,14.128 0,28.29) 16,926estem gam

TH3 : A l este bi axit no 2 chc v ancol no 2 chc Axit l

+) n =0 v n =1 ging TH1

+)n =2 th 0,65.(0,14.116 0,28.29) 15,834estem gam

Vy p n tha mn l A .

Cu 53:

Cho x gam hn hp M gm axit n chc, no (X) v ancol n chc, no (Y) c khi lng phn t bngnhau. Chia M lm 2 phn:

+ Phn 1 c khi lng 1m gam cho tc dng vi Na d thu c 420 ml kh 2 H (ktc)

+ Phn 2 c khi lng 2m gam. t chy hon ton phn 2, ly sn phm chy qua dung d ch NaOH

d, sau cho tip 2 BaCl (d) vo thu c 59,1 gam kt ta. Bit r ng t l 1 2: 1:3m m . Gi tr cax l

A. 9,0gam

B. 18,0gam

C. 12,0gam

D. 6,0gam

Li giai:

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42 T hoa h c

Gi CTPT ca X l 2 2 ( 1)n nC H O n v Y l 2 1m mC H OH

Phn 1: X Y

n a

n b

20,01875 H n mol nn 0,0375a b (1)

Phn 2:2? ?

0,3k t t a COn n mol

nn 3( ) 0,3na mb hay 0,1na mb (2)

V X Y M M nn 14 32 14 18 1n m n m

Thay vo (2) ta c: ( ) 0,1n a b b

cng vi (1) 0,10,0375

n

2n

4.60.0,0375 9 x gam

Vy p n A

Cu 54:

ho tan m gam hn hp X gm bt ca 3 oxit 2 3 Al O , ,FeO CuO c cng smol cn 240g dung

dch HCl 18,25%. Thm mt lng bt nhm cn thit vo m gam hn hp X thc hin phn ngnhit nhm, sau phn ng nhit nhm thu c cht r n Y gm 2 3 Al O , Fe v Cu . X l hn hpY

bng V ml dung dch hn hp NaOH 1M v 2( ) Ba OH 0,5M un nng sau phn ng cn 20,928 gam

cht r n khng tan. Gi tr ca V l:

A. 124ml

B. 136ml

C. 148ml

D. 160ml

Li giai:

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43 T hoa h c

Ta c2

1,2 0,6 HCl H On n

t smol mi oxit l x

Ta c 3 0,6 0,12 x x x x

0,12.56 0,12.64 14,4 20,928Fe Cum m

2 3

20,928 14, 40,064

102 Al On du

2 3 Al On phn ng vi OH l

0,60, 064 0,136

3

2 3 2 22 2 Al O OH AlO H O

0,136 .....0,272

Ta c: 0, 001 0, 5.2.0, 001 0, 272 136V V V ml

Cu 55:

Cho 100 ml dung dch 3 AgNO 1M v 3 2( )Cu NO 2M. in phn dung dich bng in cc tr vi

cng dong in 5 I A. Sau mt thi gian in phn thy khi lng dung dich giam19,6 gamng thi dung dich vn con mau xanh nhat. Cho m gam Fe vo dung dch sau in phn thu c khNO (sn phm kh duy nht) v 25,5 hn hp kim loai. Gi tr ca m l.

A.31,0 gamB.35,0 gam

C.42,0 gam

D.28,84gam

Li giai:

3 2 32( )0,2 ; 0,1Cu NO AgNOn mol n mol

(-): 22

Cu e Cu ;1 Ag e Ag ;

(+): 2 21

2 22

H O O e H

Do vn cn dung dch mu xanh nn: 2Cu d:

Ta c :2

19,6ddgiam Ag Cu Om m m m

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44 T hoa h c

219,6 0,1.108 8,8(1)Cu Om m

t 22

; 4 2 0,1(2)OCu pun x n ymol y x (BTe)

(1);(2) : 0,1 ; 0,075 x mol y mol

Dung dich sau p: 2 30,1 ; 0,4 . 0{ ^ ,_ } 5du H Cu mol n mol n NO mol

Ta c : 33 24 2Fe H NO Fe NO H

3 22Fe Fe Fe

2 2Fe Cu Fe Cu

T tnh ton tnh c 31m g

Cu 56:

4,2 gam st trong khng kh mt thi gian thu c 5,32 gam hn hp X gm st v cc oxit can. Khi cho 200 ml dung dch 3 HNO xM vo X th thy X tan ht, sinh ra 0,448 lt kh NO (ktc, sn

phm kh duy nht) v dung dch Y . Gi tr ca x l:

A.1,3

B.1,2

C.1,1

D.1,5

Li giai:

0,075.3 0,02.3 0,07.2 to ra hai mui 3 3 3 2( ) , ( )Fe NO Fe NO t smol ln lt l ,a b

Ta c:0,075 0,05

3 2 0,07.2 0,02.3 0, 2 0,025

a b a

a b b

30,05.3 0,025.2 0,02 0,22 HNOn

0, 2 0,22 1,1 x x

Cu 57:

t chy hon ton 0,08 mol hp cht hu c X thu c 21,12 gam 2CO v 7,2 gam nc. Bit

trong X cha 2 nguyn t oxi trong phn t. Cho 9,12 gam X tc dng vi dung dch KOH va

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45 T hoa h c

thu c dung Y dung dch Y tc dng vi dung dch 2 Ag O thu c 34,56 gam Ag . S ng phn

cu to ca X l.

A. 3.

B. 4

C. 5.

D. 6.

Li giai:

2 20,48( ); 0,4( )CO H On mol n mol

M X cha 2 nguyn t oxi trong phn t suy ra X l 6 10 2C H O

2

0,08( ) : 0,32( ) Ag OKOH

mol X ddY Ag mol

0,16( ) 2CHO X n mol n

V X t/d vi KOH nn X ko th ng thi cha 2 nhm CHO

X : 3 7 HCOO CH CH C H : c 3 p cu to( khng tnh p hnh hc) tha mn:

2 2 3 HCOO CH CH CH CH CH

3 3( ) HCOO CH CH CH CH CH

3 2 3( ) HCOO CH C CH CH CH

Cu 58:

Hn hp X gm Na,Mg,Cu c khi lng m c chia thnh hai phn:

-Phn 1:Cho vo nc d th thu c 2,24 lt 2 H (ktc)

-Phn 2:Cho vo HCl d th thu c 8,96 lt 2 H (ktc) v cn li 12,8 gam cht r n.

Bit 2 1 13,4m m , gi tr ca m l:

A.26,8

B.40,2

C.53,6

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46 T hoa h c

D.20,1

Li giai:

2 1 22 1

2 13,413,4

m m mm m

m m

212,8 24.0,4 12,8 23.0,8 12,8 Na Mg Cu Na Mgm m m m

222,4 31,2 31,4 49 40,2m m m

Cu 59:

Este X no, n chc, mch h, khng c phn ng trng bc. t chy 0,1mol X r i cho sn phmchy hp thhon ton vo dung d ch nc vi trong c cha 0,22mol 2( )Ca OH th vn thu c kt

ta. Thy phn X bng dung dch NaOH thu c 2 cht hu c c snguyn t cacbon trong phnt bng nhau. Phn trm khi lng ca oxi trong X l?

A. 43,24%

B. 53,33%

C. 37,21%

D. 36,36%

Li giai:

Gi2 2: n n X C H O

t chy: 2 2 2n nC H O nCO

0,1( )........0,1 ( )mol n mol

V sn phm chy hp thhon ton vo dung d ch nc vi trong c cha 0,22mol 2( )Ca OH th vn

thu c kt ta suy ra : 0, 44 0,1 0, 22n

4,4 2, 2n

M khi thy phn X bng dung dch NaOH thu c 2 cht hu c c snguyn t cacbon trong phnt bng nhau nn ta c 4n

4 8 2: X C H O

Vy % ( ) 36,36%O X

Cu 60:

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47 T hoa h c

Hn hp X gm vinylaxetilen v hiro c t khi hi so vi 2 H l 16. un hn hp X mt thi gian thu

c 1,792 lt hn hp kh Y ( ktc). Hn hp kh Y phn ng va vi dung dch cha 25,6 gam

2 Br . Th tch khng kh (cha 20% 2O v 80% 2 N v th tch, ktc) cn dng t chy hon

ton hn hp Y l:

A.35,84

B.38,08

C.7,616

D.7,168

Li giai:

Gi s mol vinylaxetylen v hiro trong hh X ln lt l x v y ta c:

52 2 32 2 3 0(1) x y M x y x y

Ta c:2

0,08 ( )(2)Y H n x y n pu

2 20,16 3 ( )(3) Br H n x n pu

Ly (3) (2) ta c: 2 0,08(4) x y

Gii hpt (3) v (4) ta c0,06

( )0,04

xmol

y

t chy Y cng nh t chy X

0,045.(0,06.5 ).22,4 35,84( )

2kk V l

Cu 61:

Cho 10,88 gam , , ,Cu Fe FeS CuS , 2Cu S , 2FeS vo dung dch cha 1,36 mol 3 HNO thu c

19,712 lt kh 2 NO duy nht v dung dch Y. Dung dch Y tc dng ti a vi m gam Cu to kh NO duy

nht. Tm m? Bit Cu, 2Cu S , CuS c smol bng nhau v Fe, 2FeS , FeS c t l mol tng ng 1:2:3.

Li giai:

Ta c:2

0,88 NOn mol

Quy i hn hp cho gm: Fe (6a mol) ;Cu (4b mol) ; S (7a + 2b)

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48 T hoa h c

33-Fe e Fe

6a---18a

22-Cu e Cu

4b----8b

24-6S e SO

(7a+2b)

3 2 NO e NO

60 20 0,88a b ; 560 320 10,88 0, 008a b a ; 0,02b

Dung dch Y cha:2

Cu (0,08mol) ;3

Fe (0,048mol); H ;2

4SO (0,096mol) v 3 NO (0,48mol)

0,192 0,48 0,16 0,144 0,368 H

n mol

22-Cu e Cu

3 2Fe e Fe

3 24 3 2 NO H e NO H O

---------0,368-----0,276

0,162 10, 368Cun m gam

Cu 62:

Cho phn ng gia 24,71 (gam) mt mui clorua ca mt nguyn tphn nhm chnh vi 10,9 (gam)amoniac to ra mt hn hp gm: 25,68 (gam) 4 NH Cl ; 2,57 (gam) mt nguyn t A th r n v 7,37

gam mui nitrua kt tinh mu vng ca A . Bit phn ng xy ra theo phng trnh:

3 4w x y t nA Cl mNH pNH Cl qA rA N

Bit r ng mt mu mui nitrua ni trn nmnh khi p bng ba, nhng khi polime ha th to ra mtcht r n dng si c mu hng , v c kh nng dn in nh kim loi. Tng gi tr ti gin ca n, w, x,m, p, q, r, y, t l.

A. 37

B. 38

C. 40

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49 T hoa h c

D. 35

Li giai:

Clm c trong 25,86g 4 NH Cl l: 35,5* 25,86 / 53,5 17,04

Am trong hp cht ban u l: 24,71 17,04 7,67

Am c trong hp cht nitrua l: 7,67 2,57 5,1 gam

N m trong hp cht nitrua l: 7,37 5,1 2,27

4ionNH m l: 25,86 17,04 8,82

N m c trong ion 4 NH = 14*8,82/18 = 6,86g

Vi w x A Cl : M l nguyn t lng ca A:

/ A C m m l = * / 35,5* M w x = 7,67 /17,04

ionCl lun lun c ha tr 1 1w 15,98* M x (1) ,x c thc cc gi tr : 1,2,3,4

Nu 1 x 15,98 M A l O loi v A th r n

Nu 2 x 31,96 M A l S tha mn 2SCl

Nu 3 x 47,94 M Loi ngay

Nu 4 x 63,92 M A l Cu loi v Cu thuc phn nhm ph.

Vi mui y t S N

/ S N m m = 32* /14 y t = 5,1/ 2,27 y t

CT ca nitrua c th l: SN , 2 2S N ....vi cc tnh cht cho CT ng: 2 2S N

Smol cc cht: 2SCl =0,24mol, 3 NH =0,64mol, 4 NH Cl =0,48mol, S =0,08mol v 2 2S N =0,08mol

n:m:p:q:r=0,24:0,64:0,48:0,08:0,08=3:8:6:1:1

23SCl + 38 NH 46 NH Cl + S + 2 2S N .

Cu 63:

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50 T hoa h c

Ha tan ht 7,20 gam hn hp A cha Al v Fe bng 800 ml dung dch 3 HNO 0,85M c dung

dch X v V lt kh NO (ktc). Thm 400 ml dung dch NaOH 1,5 M vo dung dch A thu c 67,2ml kh (ktc) c mi khai thot ra v kt ta Y . Lc bkt ta Y , c cn dung dch nc lc r i nungn khi lng khng i thu c 42,295 gam cht r n. Gi tr ca V l.

A. 3,6288 lt

B. 2,9568 lt

C. 3,1808 lt

D. 3,8528 lt

Li giai:

33 Al e Al

x--3x--x33Fe e Fe

y--3y--y

3 24 3 2 NO H e NO H O

3 4 210 8 3 NO H e NH H O

----0,03--0,024--0,003

H n d 0, 68 (4 4 0,032) 0, 03 [0,682 (4 4 )] x y x y mol

2 H OH H O

4 3 2 NH OH NH H O

333 ( )Fe OH Fe OH

333 ( ) Al OH Al OH

OH n p 0,682 (4 4 ) 0,003 3 3 [0,685 ( )] x y x y x y mol

3 4( ) [ ( ) ] Al OH OH Al OH

( * )

OH n d 0,6 0,685 ( ) ( 0,085) x y x y mol

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51 T hoa h c

Dung dch sau p: (0,6 ) Na mol ; 2 ( ) AlO xmol ; 3( 0,085);OH y NO

30,6 0,085 [0,685 ( )]

NOn x y x y mol

23*0, 6 59 17( 0, 085) 46(0,685 ) 42, 295 x y x y

29 13 1,57 y x

M ta c: 27 56 7,2 0,08 x y x ; 0,09 y0,008 0,162 3,6288 NOn x y mol V l

( * ) Ta ddng chng minh c OH d.

Cu 64:

Hn hp X gm 3 axit n chc mch h trong c hai axit no l ng ng k tip v mt axit khngno c mt lin kt i. Chom gam X tc dng vi dung dch cha 0,7 mol NaOH . trung halng NaOH d cn 200ml dung dch HCl 1M v thu c dung dch D . C cn cn thn D thuc 52,58 gam cht r n khan E . t chy hon ton E r i hp th ton b sn phm kh v hi vobnh ng dung dch d thy khi lng bnh tng 44,14 gam. Thnh phn % khi lng axit khng nol:

A. 44,89

B. 48,19

C. 40,57

D. 36,28Li giai:

t cng thc ca 2 axit no ng ng v axit khng no c 1 lin kt i ln lt l 2 2n nC H O v

2 2 2m mC H O

t smol ln lt l a,b

Tc dng va vi 0,7 0,2 0,5 mol NaOH

0,5a b (1)

55,28 0,7.18 0,7.40 0,2.36,5 29,88axit m

(14 32) (14 30) 29,88(2)a n b m

t chy hn hp cho sn phm kh v hi vo NaOH thy bnh tng 44,14 gam

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9(2 2 3 ) 44( 0, 25) 44,14an a bm b an bm (3)

T (1)(2) v (3)

0,3

0, 21,02

a

ban bm

3 2 10, 2n m

3, 1, 4m n

3 4 20,2.72 14,4C H Om

3 4 2% 48,19%C H Om

luyỆn thỊ ĐẠi hỌc hÓa 2013

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